Newton s Third Law, Momentum, Center of Mass

Team: Newton s Third Law, Momentum, Center of Mass Newton s Third Law is a deep statement on the symmetry of interaction between any two bodies in the universe. How is the pull of the earth on the moon related to the pull of the moon on the earth? How is the impact of a bat on a ball related to the impact of the ball on the bat? Newton s Third Law is intimately related to two huge principles in physics: Conservation of Momentum and Motion of the Center of Mass. Part I. Newton s Third Law Atomic Springs When you push against a wall, you feel a force in the opposite direction. The harder you push, the harder the wall pushes back on you. The amazing fact is that the force exerted by the wall on you is exactly equal and opposite to your push. How does the wall know to push this way? In general how can inanimate objects, such as walls, floors, and tables, push and pull other objects? The answer lies in the atomic world. Solids are made of atoms held together in a lattice structure by electric forces (atomic bonds). Physicists view solids as a set of balls (atoms) connected by springs (bonds): The length of one atomic spring is about one nanometer (1 9 m). The number of atoms in a typical solid such as a book or a table is about 1,,,,,,,,. When you push on a solid, the atomic springs compress, i.e., the electron clouds of the atoms overlap. The harder you push, the more the springs compress, and thus the harder the springs push back on you. What you actually feel is the electric repulsion between the atoms in the surface of the solid and the tip of your finger. It is similar to the magnetic repulsion you feel when you push one magnet toward another magnet. It is important to realize that nothing actually touches when two objects come into contact, i.e., when you push on the wall with your hand. If two atoms actually touched one nucleus on top of another nucleus then nuclear fusion would result thereby creating an atom-bomb explosion! Force fields touch, not ponderable matter. 1

This atomic springiness endows matter with an elastic property. When you put a book on the table or push on the wall, the surface of the table and the wall bend slightly (greatly exaggerated in the picture below). Think of putting a bowling ball on a mattress (box springs). Force on Table due to Book F TB F BT Force on Book due to Table The bent table acts like a compressed spring that exerts a force back on the book. The action of pushing downward creates a reaction of pushing upward. This is the atomic origin of the ubiquitous Normal Force. The Law of Action/Reaction Qualitatively speaking, a force is a push or a pull. Rigorously speaking, a force is an interaction between two objects. This interaction obeys a deep law of mechanics: Newton s Third Law: If one object exerts a force on a second object, then the second object exerts an equal and opposite force on the first. In symbols, F 12 = F 21 F 12 Force on object 1 due to object 2. F 21 Force on object 2 due to object 1. The minus sign in Newton s Third Law is one of the most important minus signs in all science. All forces come in pairs. For every action, there is an equal and opposite reaction. You cannot have a single force in any situation because the force on one object is always due to some other object. Here are some examples of interactions (pairs of forces): Action Reaction Foot push backward on Floor Floor pushes forward on Foot (explains how a person moves) Tire push backward on Road Road pushes forward on Tire (explains how a car moves) Rocket pushes back on Gas Gas pushes forward on Rocket (explains how a jet moves) Book pushes down on Table Table pushes up on Book Earth pulls down on Moon Moon pulls up on Earth Magnet pulls left on Nail Nail pulls right on Magnet Add three more force pairs to this list: 2

Part II. Experimental Tests of Newton s Third Law First make sure the track is level using the steel ball test. Here you will use two motion sensors interfaced to two computers. Sensor 1 will measure the velocity and acceleration of cart 1. Sensor 2 will measure the velocity and acceleration of cart 2. Tilt open each sensor and place at each end of the track as shown below. Sensor 1 1 2 Track Table Sensor 2 Open the program Logger Pro. Open the file Sensor 1 on the computer connected to sensor 1. Open the file Sensor 2 on the computer connected to sensor 2. Experiment 1. Magnetic Interaction (Equal Mass Carts) Make sure the cart magnets are facing each so that the action-reaction force between the carts is a force of repulsion. Give cart 1 an initial velocity (3 5 cm/s) toward cart 2 (initially at rest). Record the motion of each cart. 1 v magnetic interaction 2 Velocity Curves Your velocity curves (v as a function of time) should have the following overall appearance: initial Velocity of Cart 1 Velocity of Cart 2 final v 1 v 2 final initial t t A velocity curve is divided into three distinct regions: 1. Constant Velocity (horizontal line) just before the interaction (initial). 2. Changing Velocity (steep sloping line) during the interaction (time interval t). 3. Constant Velocity (horizontal line) just after the interaction (final). 3

CHANGE Graph Scales: Adjust the t and v scales on your graphs so that your velocity curves resemble the pictures above. The initial and final velocity plateaus should clearly fill the whole screen. Do not display any bad data that is outside the three regions defined above. Note that the interaction time t is the interval of time during which each cart feels the force due to the other cart. During the interaction time t, F 12 causes the velocity of mass m 1 to change by an amount v 1. F 21 causes the velocity of mass m 2 to change by an amount v 2. PRINT your graphs. WRITE the numerical values of t, v 1, v 2 on the corresponding parts of your velocity curves. Momentum (gain / loss) & Force (action / reaction). From your measured values of the mass and the velocity change, calculate the momentum change of each cart during the interaction. m 1 v 1 = ( kg) ( m/s) = kgm/s. m 2 v 2 = ( kg) ( m/s) = kgm/s. Compare the momentum lost by one cart and the momentum gained by the other cart. Calculate the average force F 12 on cart 1 (due to cart 2) and the average force F 21 on cart 2 (due to cart 1) during the time of interaction. Hint: Use Newton s Second Law. Be careful with signs. Show your calculation in the space below: F 12 = N. F 21 = N. CONCLUSION: 4

Force Curves During the interaction, the magnetic force on each cart varies with time. This time-dependent force function starts at zero (when the carts are far apart), then rapidly rises (as the carts get closer), then reaches a maximum (when the carts almost touch), and then falls back to zero (as the carts move farther apart). The force on cart 2 as a function of time has the following overall shape: F max F = F = t Note that the force curve is characterized by two parameters (height and width): the maximum force F max and the interaction time t. The force F max occurs when the carts are closest to each other with their magnets repelling most strongly. As always, t is the time interval during which one cart feels the force due to the other cart. What is the exact shape of this force curve? Since F is equal to ma, and m is a constant, the shape of the F curve is identical to the shape of the a curve, i.e. a 1 = F 12 /m and a 2 = F 21 /m. CHANGE the time scale on each a graph to match the time scale on the corresponding v graph. Summarize the deep (symmetrical) relationship between the overall shapes of your a 1 and a 2 curves: Calculate the maximum forces (action and reaction) during the magnetic interaction: F 12 max = m 1 a 1 max = ( ) ( ) = N. F 21 max = m 2 a 2 max = ( ) ( ) = N. Sketch your force curves below. Write the numerical values of the height and width on each curve. F 12 F 21 time time 5

Experiment 2. Magnetic Interaction (Unequal Masses) Make cart 2 more massive than cart 1 by placing the bar weight on top of cart 2. Give cart 1 an initial velocity toward cart 2 (initially at rest). Record the motion of each cart. v 1 magnet magnet 2 PRINT your graphs. Write the numerical values of the initial velocity, the final velocity, and the change in velocity of each cart directly on the corresponding parts of your velocity curves. Momentum Balance Momentum Change of Cart 1: m 1 v 1 = ( kg ) ( m/s ) = kg m/s. Momentum Change of Cart 2: m 2 v 2 = ( kg ) ( m/s ) = kg m/s. Force Curves Calculate the maximum forces during the magnetic interaction: F 12 max = m 1 a 1 max = ( ) ( ) = N. F 21 max = m 2 a 2 max = ( ) ( ) = N. CHANGE the t scale on each a graph to match the corresponding t scale on each v graph. Your two a graphs should exhibit Newtonian third-law symmetry. Sketch your measured force curves below. Write the numerical values of the height and width on each curve. F 12 F 21 time time Based on your F graphs, what does your experiment prove? 6

Part III. Conservation of Momentum What does Action = Reaction have to do with Momentum = constant? Everything! From your experimental tests of Newton s Third Law, you discovered that the amount of mv lost by one object is equal to the amount of mv gained by the other object. This means that the total amount of mv in the system cannot be created or destroyed it is conserved! Experiment 1. Create Velocity Reaction Here is the initial state of the system: 1 2 Push the equal-mass carts together with your hands until the magnets on each cart touch. In this initial state, the magnetic repulsion is ready to explode the system. Release both carts at the same time and observe them fly apart. Based solely on your observation (no quantitative measurements), sketch three momentum curves (momentum as a function of time): p 1 of cart 1, p 2 of cart 2, and p 1 + p 2 of the system. Label your three curves. Neglect friction. Momentum Time Experiment 2. Velocity Exchange Reaction Here is the initial state of two equal-mass carts: v 1 magnet magnet Rest 2 Based solely on observing this reaction, sketch and label p 1, p 2, and p 1 + p 2 as time flows. Momentum Time 7

Part IV. Design Project. Reduce Speed by 2/5 Here your team will study a one-dimensional, completely-inelastic collision between two carts. Before the collision, cart 1 moves with some initial speed toward cart 2. Cart 2 is initially at rest. The DESIGN SPECS require: 1. After the collision, the two carts must remain stuck together. 2. The final speed of the joined carts must equal 2/5 the initial speed of cart 1. Theory Derive the theoretical relationship between the cart masses, m 1 and m 2, that will give the 2/5 speed reduction. Work out your theory in the space below. Equipment How do you plan to get the carts to stick together? What are the masses (in kilograms) of the two carts you plan to use to test your theory? Experiment Perform the collision experiment with the two-cart system you designed. Use a sensor to record the motion. PRINT your velocity vs time graph. Change the graph scales so that the initial and final speed plateaus are clearly displayed. WRITE the values of the initial and final speeds on your graph. In the space below, compare your experimental data with the design specs. 8

Part V. The Center of Mass: A Case Study What is the Center of Mass? Place cart 1 (loaded with the bar weight) on the track so that the center of cart 1 is at x 1 = 2 cm. The x axis is defined by the meter scale fixed to the track. Place cart 2 (no added weight) so that its center is at x 2 = 8 cm. The masses of the carts are m 1 = 1. kg and m 2 =.5 kg. The center of mass x cm of this system of two carts is 1. kg 2 cm +.5 kg 8 cm x cm = = 4 cm. 1. kg +.5 kg 1. kg center of mass.5 kg 2 cm 4 cm 8 cm Discovering a Conservation Law Now attach a spring between the carts (from the right side of the 1. kg mass to the right side of the.5 kg mass). Make sure the Velcro patches are facing each other so that the carts stick together when they collide. Pull the carts apart so that their centers are at 2 cm and 8 cm as before. This is the initial state. Release the carts at the same time. Observe the motion. The carts accelerate toward each other, collide, and come to rest (final state). Record the final resting positions (x 1 and x 2 ) of the center of cart 1 and the center of cart 2 in the table below. Calculate the center of mass x cm of the system of two carts in this final state. SHOW your calculation in the space below: m 1 m 2 x 1 x 2 x cm initial 1. kg.5 kg 2 cm 8 cm 4 cm final 1. kg.5 kg Does your experimental data prove that the center of mass is conserved? equal x cm (final) within experimental error (1%)? i.e. Does x cm (initial) 9

The Center of Mass is the Place to Be Why is x cm conserved? Because F 12 = F 21! Here is the Newtonian logic: F 12 + F 21 = implies m 1 a 1 + m 2 a 2 = which says a cm =. Amidst the chaos of accelerating parts (the carts rushing toward each other), there is one point of calmness : the center of mass x cm. i.e. x 1 and x 2 are complicated (depend on time t). m 1 x 1 +m 2 x 2 is simple (independent of t). The conservation relation x cm = constant allows you to know the position of each cart at any time during the motion without having to know anything about the time-dependent internal (spring) force that is causing the motion! The initial position 2 cm of cart 1 (1. kg) and the initial position 8 cm of cart 2 (.5 kg) are marked on the position diagram below. The position of cart 1 is also marked at three later times of the motion. Mark the position of cart 2 at these same three future times. Mark the position of the center of mass of the system at all four times with the symbol. 1. kg.5 kg t o 2 cm 3 cm 4 cm 5 cm 6 cm 7 cm 8 cm t 1 25 cm t 2 3 cm t 3 35 cm Einstein, Center of Mass, E = mc 2 The two Conservation Laws that you discovered in this laboratory, m 1 v 1 + m 2 v 2 = constant and m 1 x 1 + m 2 x 2 = constant, are two of the most important consequences of Newton s Third Law F 12 + F 21 =. Albert Einstein realized the deep significance of the constancy of x cm. In his 196 paper, The Principle of Conservation of Motion of the Center of Mass and the Inertia of Energy, Einstein used the relation x cm (initial) = x cm (final) to derive the most famous equation in science: E = mc 2. Einstein used this center of mass theorem to analyze the motion of a train car when a light bulb inside the car was turned on. The light source emits photons to the right which causes the train to recoil to the left. The center of mass does not move! 1